1. A particle moves for 4 seconds in a straight line with uniform acceleration and describes 52 m. It then travels with uniform speed covering 48 m in 3 s. It is brought to rest by a retardation twice that of the initial acceleration.

To get the initial velocity and acceleration of the particle... well, we can't. There isn't enough information. I am going to assume that the motion is "smooth:" the final velocity of the initial phase is the same as the constant velocity of the second phase.

So assuming an origin at the point where the particle was when the time started and a +x direction in the direction of the initial acceleration we have:

where

So

Solving the top equation for v_0 I get:

Inserting this into the second equation gives:

Solving the quadratic for a I get that: or

We discard the negative solution since the acceleration is defined to be positive in the first phase, so .

Thus

The final retardation (acceleration) is just twice the negative of the acceleration in the first part so .

You can use to find the distance travelled in the last phase of the motion. I leave that to you.

thanks alot

thank you Dan , i tried opening it and it is readable ,i was wondering why it didnt open with u ,i think u just need to point the curser on it and maximize it i did that and it was readable ...thanks again Dan !!